If for a non - zero x, 3x² + 5x + 3 = 0, then the value of $${x^3} + \frac{1}{{{x^3}}}$$   is?

$$\eqalign{ & 3{x^2} + 5x + 3 = 0 \cr & 3{x^2} + 3 = - 5x \cr & {\text{Divide by }}3x{\text{ both sides}} \cr & x + \frac{1}{x} = \frac{{ - 5}}{3} \cr & {\text{Then, }}{x^3} + \frac{1}{{{x^3}}} \cr & = {\left( {\frac{{ - 5}}{3}} \right)^3} - 3 \times \left( {\frac{{ - 5}}{3}} \right) \cr & = \frac{{ - 125}}{{27}} + 5 \cr & = \frac{{10}}{{27}} \cr} $$